# Packet Substitution Solutions - Integration by Substitution...

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Integration by Substitution SolutionsNote:in (1) through (6), we find the answer by settingduequal to thederivative of the function, multiplied bydx.1.u= 4x+ 1du= 4dx2.u=x4+ 2du= 4x3dx3.u= 3x2+ 1du= 6xdx4.u= ln(x)du=1xdx5.u= 1/x=x-1du=-x-2dx6.u=ex+ 2du=exdx7.u= 4x+ 1du= 4dxdu/4 =dxR(4x+ 1)4dx=Ru4du4=14Ru4du=14·u55+C=120(4x+ 1)5+C8.u=-2x+ 3du=-2dx-du/2 =dxRe-2x+3dx=Reu· -du2=-12Reudu=-12eu+C=-12e-2x+3+C9.Note:in problems like this one, it may not be immediately obviouswhat to setuequal to (e.g., here, you might setu= 5x3oru=x4+2). Youcan try trial and error, by settinguequal to something and trying to solve1
2the problem, and then starting over if you can’t cancel out all of the ‘x’s;another tip is to look for a function whose derivative you see in the integral.So if you setu=x4+ 2, the derivative is 4x3, which is similar to the 5x3we already see in the integral; this is a good indicator thatu=x4+ 2 is theright choice.

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Term
Spring
Professor
N/A
Tags
Berlin U Bahn, dx, Heinrich Lehmann Willenbrock
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